# Definition for a binary tree node.
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution:
    def sortedArrayToBST(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        def to_bst(nums, start, end):
            if start > end:
                return None
            mid = (start + end) // 2
            node = TreeNode(nums[mid])
            node.left = to_bst(nums, start, mid - 1)
            node.right = to_bst(nums, mid + 1, end)
            return node

        return to_bst(nums, 0, len(nums) - 1)

if __name__ == '__main__':
    nums = [-10,-3,0,5,9]






    """
            Time Complexity = O(n)
            Space Complexity = O(log(n))

            Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
            For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of
            the two subtrees of every node never differ by more than 1.

            Example:
            Given the sorted array: [-10,-3,0,5,9],
            One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
                  0
                 / \
               -3   9
               /   /
             -10  5
    """
